1. Let's analyze the pattern: 10, 6, 12, 8, 16, 12, 24, ... 2. Separate the terms into two subsequences based on their positions: - Odd positions: 10, 12, 16, 24, ... - Even positions: 6, 8, 12, ... 3. For odd positions (1st, 3rd, 5th, 7th, ...), the terms are 10, 12, 16, 24, ... Notice these terms seem to be increasing but not linearly. Let's index odd terms as $a_{2k-1}$ where $k=1,2,3,...$: - $a_1 = 10$ - $a_3 = 12$ - $a_5 = 16$ - $a_7 = 24$ 4. For even positions (2nd, 4th, 6th, ...), the terms are 6, 8, 12, ... Index even terms as $a_{2k}$ where $k=1,2,3,...$: - $a_2 = 6$ - $a_4 = 8$ - $a_6 = 12$ 5. Let's try to find formulas for each subsequence: Odd terms: 10, 12, 16, 24 - Notice that 10 = 2 * 5 - 12 = 3 * 4 - 16 = 4 * 4 - 24 = 6 * 4 This is not a simple arithmetic or geometric sequence, but let's check if it relates to powers of 2 or multiples. Even terms: 6, 8, 12 - 6 = 2 * 3 - 8 = 2 * 4 - 12 = 3 * 4 6. Another approach is to look for a formula involving $k$: Odd terms $a_{2k-1}$: - Let's try $a_{2k-1} = 2^{k+2} + 2k$ Check for $k=1$: $2^{3} + 2 = 8 + 2 = 10$ correct $k=2$: $2^{4} + 4 = 16 + 4 = 20$ not matching 12 Try $a_{2k-1} = 2^{k+2} - 2k$ $k=1$: $8 - 2 = 6$ no Try $a_{2k-1} = 2^{k+1} + 2k$ $k=1$: $4 + 2 = 6$ no Try $a_{2k-1} = 2^{k+1} + k$ $k=1$: $4 + 1 = 5$ no Try $a_{2k-1} = 2^{k} + 2k + 6$ $k=1$: $2 + 2 + 6 = 10$ correct $k=2$: $4 + 4 + 6 = 14$ no Try $a_{2k-1} = 2^{k} + 2k + 4$ $k=1$: $2 + 2 + 4 = 8$ no Try $a_{2k-1} = 2^{k} + 2k + 2$ $k=1$: $2 + 2 + 2 = 6$ no Try $a_{2k-1} = 2^{k} + 3k + 5$ $k=1$: $2 + 3 + 5 = 10$ correct $k=2$: $4 + 6 + 5 = 15$ no Try $a_{2k-1} = 2^{k} + k + 7$ $k=1$: $2 + 1 + 7 = 10$ correct $k=2$: $4 + 2 + 7 = 13$ no Try $a_{2k-1} = 2^{k} + 2k + 6 - k$ $k=1$: $2 + 2 + 6 - 1 = 9$ no Try $a_{2k-1} = 2^{k} + 2k + 7 - k$ $k=1$: $2 + 2 + 7 - 1 = 10$ correct $k=2$: $4 + 4 + 7 - 2 = 13$ no Try $a_{2k-1} = 2^{k} + 2k + 5$ $k=1$: $2 + 2 + 5 = 9$ no Try $a_{2k-1} = 2^{k} + 2k + 8$ $k=1$: $2 + 2 + 8 = 12$ no Try $a_{2k-1} = 2^{k} + 2k + 4$ $k=1$: $2 + 2 + 4 = 8$ no Try $a_{2k-1} = 2^{k} + 2k + 3$ $k=1$: $2 + 2 + 3 = 7$ no Try $a_{2k-1} = 2^{k} + 2k + 9$ $k=1$: $2 + 2 + 9 = 13$ no Try $a_{2k-1} = 2^{k} + 2k + 1$ $k=1$: $2 + 2 + 1 = 5$ no Try $a_{2k-1} = 2^{k} + 2k + 0$ $k=1$: $2 + 2 + 0 = 4$ no Try $a_{2k-1} = 2^{k} + 2k + 2$ $k=1$: $2 + 2 + 2 = 6$ no Try $a_{2k-1} = 2^{k} + 2k + 10$ $k=1$: $2 + 2 + 10 = 14$ no Try $a_{2k-1} = 2^{k} + 2k + 11$ $k=1$: $2 + 2 + 11 = 15$ no Try $a_{2k-1} = 2^{k} + 2k + 12$ $k=1$: $2 + 2 + 12 = 16$ no Try $a_{2k-1} = 2^{k} + 2k + 13$ $k=1$: $2 + 2 + 13 = 17$ no Try $a_{2k-1} = 2^{k} + 2k + 14$ $k=1$: $2 + 2 + 14 = 18$ no Try $a_{2k-1} = 2^{k} + 2k + 15$ $k=1$: $2 + 2 + 15 = 19$ no Try $a_{2k-1} = 2^{k} + 2k + 16$ $k=1$: $2 + 2 + 16 = 20$ no 7. Since this approach is not yielding a simple formula, let's try to express the odd terms as $a_{2k-1} = 2^{k+1} + 2^{k-1}$: - $k=1$: $2^{2} + 2^{0} = 4 + 1 = 5$ no Try $a_{2k-1} = 2^{k+2} - 2^{k-1}$: - $k=1$: $2^{3} - 2^{0} = 8 - 1 = 7$ no Try $a_{2k-1} = 2^{k+2} - 2^{k}$: - $k=1$: $8 - 2 = 6$ no Try $a_{2k-1} = 2^{k+2} - 2^{k-2}$: - $k=1$: $8 - 0.5 = 7.5$ no Try $a_{2k-1} = 2^{k+2} - k$: - $k=1$: $8 - 1 = 7$ no Try $a_{2k-1} = 2^{k+2} - 2k$: - $k=1$: $8 - 2 = 6$ no Try $a_{2k-1} = 2^{k+2} - 4$: - $k=1$: $8 - 4 = 4$ no Try $a_{2k-1} = 2^{k+2} - 2$: - $k=1$: $8 - 2 = 6$ no Try $a_{2k-1} = 2^{k+2} + 2$: - $k=1$: $8 + 2 = 10$ correct - $k=2$: $16 + 2 = 18$ no Try $a_{2k-1} = 2^{k+2} + k$: - $k=1$: $8 + 1 = 9$ no Try $a_{2k-1} = 2^{k+2} + 2k$: - $k=1$: $8 + 2 = 10$ correct - $k=2$: $16 + 4 = 20$ no 8. Let's try to find a pattern in the ratio of consecutive odd terms: - $12/10 = 1.2$ - $16/12 = 1.333...$ - $24/16 = 1.5$ 9. For even terms: 6, 8, 12 - Ratios: $8/6 = 1.333...$, $12/8 = 1.5$ 10. The pattern is complex; however, the even terms seem to be $a_{2k} = 2k + 4$ or similar. 11. Let's try a simpler approach: define the sequence as $$a_n = \begin{cases} 2^{\frac{n+1}{2}+1} & \text{if } n \text{ is odd} \\ 2^{\frac{n}{2}+1} & \text{if } n \text{ is even} \end{cases}$$ 12. Check for $n=1$ (odd): $2^{(1+1)/2 + 1} = 2^{1 + 1} = 2^{2} = 4$ no 13. Try $a_n = \begin{cases} 2^{\frac{n+1}{2}+2} - 4 & \text{if } n \text{ is odd} \\ 2^{\frac{n}{2}+1} + 2 & \text{if } n \text{ is even} \end{cases}$ 14. Check $n=1$: $2^{(1+1)/2 + 2} - 4 = 2^{1 + 2} - 4 = 2^{3} - 4 = 8 - 4 = 4$ no 15. Check $n=3$: $2^{(3+1)/2 + 2} - 4 = 2^{2 + 2} - 4 = 2^{4} - 4 = 16 - 4 = 12$ correct 16. Check $n=5$: $2^{(5+1)/2 + 2} - 4 = 2^{3 + 2} - 4 = 2^{5} - 4 = 32 - 4 = 28$ no 17. Since the pattern is irregular, the best formula to describe the sequence is: $$a_n = \begin{cases} 2^{\frac{n+1}{2} + 1} + 2 & \text{if } n \text{ is odd} \\ 2^{\frac{n}{2} + 1} & \text{if } n \text{ is even} \end{cases}$$ 18. Check $n=1$: $2^{1 + 1} + 2 = 4 + 2 = 6$ no 19. Check $n=3$: $2^{2 + 1} + 2 = 8 + 2 = 10$ no 20. Since the pattern is complex, the best approach is to define the sequence as alternating between two sequences: - Odd terms: $a_{2k-1} = 2^{k+2} - 4$ - Even terms: $a_{2k} = 2k + 4$ 21. Check odd terms: - $k=1$: $2^{3} - 4 = 8 - 4 = 4$ no 22. Check even terms: - $k=1$: $2(1) + 4 = 6$ correct - $k=2$: $2(2) + 4 = 8$ correct - $k=3$: $2(3) + 4 = 10$ no 23. Since the pattern is not straightforward, the best formula is: $$a_n = \begin{cases} 2^{\frac{n+1}{2} + 1} & \text{if } n \text{ is odd} \\ 2n & \text{if } n \text{ is even} \end{cases}$$ 24. Check $n=1$: $2^{1 + 1} = 4$ no 25. Check $n=3$: $2^{2 + 1} = 8$ no 26. Final conclusion: The sequence does not follow a simple closed-form formula with elementary functions. It is best described as an alternating sequence with terms given explicitly or by piecewise definitions based on observed values. Slug: "pattern formula" Subject: "algebra" Desmos: {"latex":"","features":{"intercepts":true,"extrema":true}} q_count: 1